14.126 Game Theory Final Exam Youhave24hoursf romt hetimeyoup ick upt heexam( i.e. youneedtoreturn your solutions by the same time next day). You can use any existing written source, but you cannot discuss the content of this exam with others. Questions are equally weighted. Good Luck. 1. This question tries to evaluate your understanding of the basic solution concepts. (a) (10 points) Consider a class of games Gθ =(N,S,u( ; θ)),indexedb ya p ayoff parameter θ which comes from a compact metric space. ·Assume that player set N and strategy space S are finite. State broad conditions on u under which whenever Gθ∗ has a unique Nash equilibrium s∗ in pure strategies, θ∗ has a neighborhood η θ such that for each θ η, Gθ has a unique Nash equilibrium s in pure strategies θ ∈ and s = s∗. Answer: Assume that u is continuous in θ.SinceS is finite, u is also continuous in s. Then, as proved in the lecture slides, the correspondence PNE of Nash equilibria in pure strategies is upper-hemicontinuous and non-empty. Since S is finite, this means that there exists a neighborhood η of θ∗ such that for each θ η, ∈ PNE(θ) PNE(θ∗).SincePN E(θ) = ∅ and PNE(θ∗)= s∗ , this implies ⊆ 6 { } that PNE(θ)= s∗ for each θ η. { } ∈ (b) (15 points) Compute the sequential equilibria of the following Bayesian game. There are two players and a payoff parameter θ 0, 1 . Player 1 knows the value of θ, and player 2 knows the value of a signal x∈ { 0,}1 . The joint distribution of θ and x is given by ∈ { } Pr ((θ, x)=(0, 0)) = Pr ((θ, x)=(1, 0)) = Pr ((θ, x)=(1, 1)) = 1/3. The payoffsanda ctionsare asi nt hefollowing" tree": 1 R 2 b 1 r 3θ,3 L a l 2,2 0,2 3(1-θ),0 Answer: The incomplete information game is as follows, where I also indicated 1 the necessary moves in any sequential equilibrium, the moves that are obvious: 1 R 2 b 1 r 3,3 L a l θ=x=1 2,2 0,2 0,0 1/3 θ=1 1 R 2 b 1 r x=0 3,3 μ 1/3 L a l 1/3 2,2 0,2 0,0 θ=x=0 1 R 2 b 1 r 0,3 L a l 2,2 0,2 3,0 Write μ =Pr(θ =1x =0,R) for the probability player 2 assigns on her nontrivial information set. Write| β for the probability that she plays b at that information set. Write p =Pr(R θ =1) for the probability player 1 plays R on the upper branches. Similarly, w|rite q =Pr(R θ =0). Notice that player 1 assigns 1/2 to x =1on his first non-trivial informat|ion set (θ =1;emptyh istory). Now, at that information set, if player 1 plays R, he gets 3 with probability 1/2+(1/2) β =(1+β) /2 and 0 with the remaining probability. He gets 2 from playing L, so by sequential rationality 1 if β>1/3 p = [0, 1] if β =1/3 ⎧ ⎨ 0 if β<1/3. When θ =0,hegets3 with prob⎩ability β if he plays R and2f orsureifheplays L. Hence, by sequential rationality 1 if β>2/3 q = [0, 1] if β =2/3 ⎧ ⎨ 0 if β<2/3. Finally sequential rationality for⎩ player 2 implies that 1 if μ>2/3 β = [0, 1] if μ =2/3 ⎧ ⎨ 0 if μ<2/3. Towards computing sequential eq⎩uilibria, let’s rule out certain values for β. Firstly, if β>2/3,thenp = q =1, and hence by consistency μ =1/2 < 2/3,showingt hat β =0, a contradiction. Hence, β 2/3. On the other hand, if β (1/3, 2/3), then p =1and q =0,sot hatμ =0≤ (by consistency), yielding β =0∈ ,another contradiction. Hence, β [0, 1/3] 2/3 . These values indeed correspond to the two components of seque∈ntial equ∪ilibria{ .} 2 First component: β =2/3; μ =2/3 (by sequential rationality of 2); p =1,andq =1/2 (by consistency).1 Second Component: p = q =0; β 2/3; μ 2/3. ≤ ≤ Given the above equations, it is easy to verify that each combination in the above com- ponents yields a sequential equilibrium. There is a continuum of sequential equilibria, but there are only two sequential equilibrium outcomes. 2. Consider an infinitely repeated game with discount factor δ and the stage game C D C 5, 5 0, 5+θ D 5+θ, 0 θ, θ where θ ¯θ, ¯θ for some ¯θ 5. All previous actions are common knowledge. ∈ − À (a) (7 poi£nts) A¤ssuming that θ is common knowledge and fixed throughout the re- peated game, compute the best "strongly symmetric" subgame perfect equilib- rium. (A strategy profile is strongly symmetric if the players’ actions are identical for every given history.) Answer: When θ 0, the best equilibrium is clearly, each player playing their dominant dominan≤t C at all histories. Similarly, when θ 5, always playing the dominant action D is the best SPE (the best possible outcom≥ e). Now consider the case θ (0, 5). As in Abreu’s theorem, we can take the outcome path after a deviation a∈s the worst SPE, which is playing the static equilibrium (D, D) forever. On the path they play (C, C). In order this to be an equilibrium it must be that 5 (5 + θ)(1 δ)+δθ, i.e., ≥ − 5δ θ. ≥ Hence, when θ 5δ, the the best strongly symmetric SPE is play (C, C) until someone deviates,≤ and play (D, D) thereafter. It is always play (D, D) when θ>5 δ. (b) (7 points) Assume that ev ery day a new θ is drawn, and (θ0,θ1,...) are iid with ¯ ¯ uniform distribution on θ, θ ,whereθ t is the θ at t.Assumea lsot hatθ t becomes common knowledge at t.− Consider the following Grim Trigger strategy: on the £ˆ ¤ path play C at t iff θt θ, and if any player deviates play C iff θt 0 thereafter. [The cutoff is ˆθ on the ≤path and 0 off the path.] Find the conditions≤ (on θˆ and δ) under which each player playing the Grim Trigger strategy is a subgame-perfect equilibrium. Answer: The value of being on the path is 1 θˆ θˆ + ¯θ θˆ f ˆθ = − + 5. 2¯θ · 2 2θ¯ · ³ ´ The value of deviation is f (0). For this to be an equiliubrium, for each θ θˆ,we ≤ must have 5(1 δ)+δf ˆθ (5 + θ)(1 δ)+δf (0) − ≥ − ³ ´ 1 μ = p/ (p + q), i.e., 2/3=1/ (1 + q) . 3 i.e., δ θ f θˆ f (0) . (1) ≤ 1 δ − − Therefore, we must have h ³ ´ i δ θˆ f θˆ f (0) . ≤ 1 δ − − h ³ ´ i In order for playing D when θ>ˆθ ,its uffices that θˆ 0. For the best equilibrium, we consider the largest ˆθ with equality. For convenien≥ce, let’s rewrite the equation as ˆ f θ f (0) 1 δ g θˆ − = − . (2) ≡ ³ ´ θˆ δ ³ ´ Since f is concave, g is decreasing in ˆθ, and hence there is a unique solution to this equation, which you can compute algebraically. Using (1) and the single deviation principle, one can easily show that this condition is also sufficient. (c) (7 points) In part (b), assume instead that the players do not know θt,bute ach player i observes a signal xi,t = θt + εη at date t,whereε> 0 is small, the i,t random variables θt,η1,t,η2,t are all stochastically independent, and each ηi,t is uniformly distributed on [ 1, 1]. Consider the following Grim Trigger strategy: ¡ − ¢ At the initial state, play C at t iff x i,t ˆx. If any player plays D, switch to the ≤ punisment state, and remain there forever. At the punishment state, play C iff x i,t 0. ≤ Find the conditions under which each player playing the Grim Trigger strategy is a perfect Bayesian Nash equilibrium in the limit ε 0. → Answer: (This question was inspired by Sylvain Chassang’s dissertation.) In the limit ε 0, the value of being at the initial state is F (ˆx) and the value of being in the puni→ shment state is f (0) (asinp art(b)). Thevalue F (ˆx) is different from f (ˆx) becaus e of p ossibility of punisment on the path. To compute F, letting π = xˆ + ¯θ / 2θ¯ to be the probability of remaining on the path, we write ¡ F (ˆ¢x)=¡ π ¢ [5 (1 δ)+δF (ˆx)] + (1 π) xˆ + ¯θ / 2θ¯ + δf (0) · − − and obtain £¡ ¢ ¡ ¢ ¤ π5(1 δ)+( 1 π) xˆ + ¯θ / 2θ¯ +(1 π) δf (0) F (ˆx)= − − − (3) 1 δπ ¡ ¢ ¡ ¢ f (ˆx)+( 1 π) δf (0) − = − . 1 δπ − Hence, the augmented game in the limit is C D C 5(1 δ)+δF (ˆx) , 5(1 δ)+δF (ˆx) δf (0) , (1 δ) x2 + δf (0) . − − − D (5 + xi)(1 δ)+δf (0) ,δf (0) (1 δ) xi + δf (0) , (1 δ) xj + δf (0) − − − 4 Notice that probability that the other player plays C is Pr (xj xˆ xi). Hence, one must play C (in the limit) iff ≤ | δ [F (ˆx) f (0)] Pr (xj xˆ xi) (1 δ) xi. (4) − ≤ | ≥ − By continuity (for ε>0 ), we must have δ [F (ˆx) f (0)] Pr (xj xˆ xi =ˆx)=(1 δ)ˆx, − ≤ | − i.e., δ [F (ˆx) f (0)] /2=(1 δ)ˆx, − − yielding F (ˆx) f (0) 1 δ − =2 − .